If f(x)=4 x^{3}+3 x^{2}+3 x+4 ; x \neq 0, then \frac{d}{d x}(x^{3} \cdot f(\frac{1}{x}))=

24 x^{5}+15 x^{4}+12 x^{3}+12 x^{2}

\frac{x^{2}}{12}+\frac{x}{6}+\frac{1}{3}